제출 #1135111

#제출 시각아이디문제언어결과실행 시간메모리
1135111Lemser수천개의 섬 (IOI22_islands)C++20
27.75 / 100
19 ms7496 KiB
#include "islands.h"
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;

using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;

// #define endl '\n'
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()

#define fore(i,l,r) for(auto i=l;i<r;i++)
#define fo(i,n) fore(i,0,n)
#define forex(i,r,l) for(auto i=r; i>=l;i--)
#define ffo(i,n) forex(i,n-1,0)

bool cmin(ll &a, ll b) { if(b<a){a=b;return 1;}return 0; }
bool cmax(ll &a, ll b) { if(b>a){a=b;return 1;}return 0; }
void valid(ll in) { cout<<((in)?"Yes\n":"No\n"); }
ll lcm(ll a, ll b) { return (a/gcd(a,b))*b; }
ll gauss(ll n) { return (n*(n+1))/2; }

variant<bool, vector<int>> find_journey(
    int n, int m, vector<int> u, vector<int> v) {
  if (n == 2) {
    vll ok[2];
    fo (i, m) ok[u[i]].pb(i);
    if (sz(ok[0]) < 2 || sz(ok[1]) < 1) return false;
    vector<int> r;
    r.pb(ok[0][0]);
    r.pb(ok[1][0]);
    r.pb(ok[0][1]);
    r.pb(ok[0][0]);
    r.pb(ok[1][0]);
    r.pb(ok[0][1]);
    return r;
  }
  vector<vector<array<ll, 2>>> gr(n);
  fo (i, m) {
    gr[u[i]].pb({v[i], i});
  }
  vector<int> r;
  // r.pb(ady[0][1]);
  // r.pb(ady[1][0]);
  // r.pb(ady[0][2]);
  // r.pb(ady[2][0]);
  // r.pb(ady[1][0]);
  // r.pb(ady[0][1]);
  // r.pb(ady[2][0]);
  // r.pb(ady[0][2]);
  set<ll> vis;
  ll uc = 0;
  while (1) {
    vis.insert(uc);
    ll c = 0;
    vll dx, ux;
    for (auto [vc, i]: gr[uc]) {
      if (vis.find(vc) != vis.end()) continue;
      dx.pb(i);
      ux.pb(vc);
      c++;
    }
    // cout << c << '\n';
    if (c == 0) break;
    if (c > 1) {
      auto b = r;
      reverse(all(b));
      ll e1 = dx[0], e2 = dx[1];
      ll e3 = (e1&1 ? e1-1 : e1+1);
      ll e4 = (e2&1 ? e2-1 : e2+1);
      r.pb(e1);
      r.pb(e3);
      r.pb(e2);
      r.pb(e4);
      r.pb(e3);
      r.pb(e1);
      r.pb(e4);
      r.pb(e2);
      for (ll uax: b) r.pb(uax);
      return r;
    }
    r.pb(dx[0]);
    uc = ux[0];
  }
  return false; 
}
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